Gradient of the entrywise 1-norm of a pseudoinverse matrix Given a matrix $A \in \mathbb{R}^{n×m}$, and its pseudoinverse $A^{\dagger} \in \mathbb{R}^{m×n}$, how can I calculate the following gradient: 
$\frac{\partial \Vert {A}^{\dagger} \Vert_1 }{\partial A}$
where $\Vert \cdot \Vert_1$ is the entrywise 1-norm: 
$\vert\vert{ A}\vert\vert _1=\sum_{i=1}^n\sum_{j=1}^m \vert a_{ij}\vert $ ?
I have tried searching trough identities in the differential form, such as:


*

*$d(|A|) = |A|\operatorname {tr} (A ^{-1}d{A} )$ (which seems to be only defined for $A \in \mathbb{R}^{n×n}$ ?)

*$dy=\operatorname {tr} (B \,dA ) \iff \frac{dy}{dA} = B $


and use them to derive some results, but had no success when checking my expressions against numerical gradient.
 A: In the following, the scalar function
$$\eqalign{
\operatorname{sign}(x) &= \begin{cases} +1 &\text{if }(x\ge 0) \\ -1 & \text{otherwise}\end{cases} \cr
}$$
will be applied element-wise for matrix arguments.
Starting with the well-known result for the differential of the pseudoinverse
$$\eqalign{
 G &= A^{\dagger} \cr
dG &= GG^T\,dA^T\,(I-AG) + (I-GA)\,dA^T\,G^T G - G\,dA\,G \cr 
}$$
First write the L1-norm in terms of the Frobenius product (denoted by a colon) and the sign function.
Then finding the differential and gradient is straightforward. 
$$\eqalign{
 L &= \operatorname{sign}(G):G \cr
   &= S:G \cr\cr
dL &= S:dG \cr
   &= S:(GG^T\,dA^T\,(I-AG) + (I-GA)\,dA^T\,G'G - G\,dA\,G) \cr
   &= S:GG^T\,dA^T\,(I-AG) + S:(I-GA)\,dA^T\,G^T G - S:G\,dA\,G \cr
   &= (I-AG)S^TGG^T:dA + G'GS^T(I-GA):dA - G^TSG^T:dA \cr
   &= ((I-AG)S^TGG^T + G^T GS^T(I-GA) - G^TSG^T):dA \cr
\cr
\frac{\partial L}{\partial A} &= (I-AG)S^TGG^T + G^T GS^T(I-GA) - G^TSG^T \cr\cr
}$$
If you are unfamiliar with the Frobenius product, it's an infix operator which is equivalent to the trace $$A:B = \operatorname{tr}(A^TB)$$
